Sixth Edition 3.11

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904901860
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Joined: Mon Jan 08, 2018 8:16 am

Sixth Edition 3.11

Postby 904901860 » Tue Oct 30, 2018 7:46 pm

I'm confused as to why it would be a.) Co+3 and b.) Fe3+ instead of a.) Fe+3 and b.)Mn +3

jguiman4H
Posts: 31
Joined: Fri Sep 28, 2018 12:29 am

Re: Sixth Edition 3.11

Postby jguiman4H » Tue Oct 30, 2018 8:38 pm

So for a and b of this problem, we're given: (a) [Ar]3d6; (b) [Ar]3d5 (and that these are M3+ where M is a metal).

This means that to find the neutral atom (so we know which element we're looking for) we'd need to look for the electron configuration with 3 more electrons than what we were given (accounting for the 3+ charge).

-in part a, this would be a configuration of [Ar]4s23d7. This is Co, which is why for the ionized structure [Ar]3d6 we'd get Co3+.

-in part b, this would be a configuration of [Ar]4s23d6. This is the typical configuration for Fe, which is why for the ionized structure [Ar]3d5 we'd get Fe3+.


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